After two plays the wildcats gained a total of 16 yards. If X represents the number of yards for play one, and Y represents the
number of yards for play two make a table for possible values of X and Y. Graph the ordered pairs and describe the graph
1 answer:
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Answer:
$4800
Step-by-step explanation:
The cost is minimized when any pair of opposite faces of the box costs the same as any other pair. Since the top+bottom total $25 per square foot and the sides (left+right or front+back) total $20 per square foot, the area of a side must be 25/20 = 5/4 times the area of the top or bottom.
If we let x represent the length of one side of the square bottom, then 5/4x will be the height of the box. Its volume will be ...
(5/4x)(x²) = 640 ft³
x³ = (4/5)(640 ft³) = 512 ft³ . . . . . . multiply by 4/5
x = 8 ft . . . . . . . . . . . . . . . . . . . . . . . take the cube root
The base area is then (8 ft)² = 64 ft², and the cost of the top+bottom is ...
64 ft² × $25/ft² = $1600
The three pairs of opposite sides will cost 3×$1600 = $4800.
The minimum cost of the constructed box is $4800.
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You can also write an equation for the total cost in terms of x, the side length of the base. That cost function will be ...
c(x) = 15x² + 10x² + 10·4x(640/x²)
This simplifies to ...
c(x) = 25x² +25600/x
A graph shows the minimum is c(8) = 4800 (as above).
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Algebraically, you can set the derivative to zero to find the value of x.
dc/dx = 0 = 50x -25600/x²
512 = x³ . . . . multiply by x²/50 and add 512
8 = x . . . . . . . take the square root. Same solution as above. <em>Cost is minimized when the side length of the base is 8 ft</em>.
